Conservation of energy - the elevator question

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Homework Help Overview

The discussion revolves around a physics problem involving the conservation of energy in the context of an elevator that falls onto a cushioning spring after its cable snaps. The problem includes calculations related to the speed of the elevator just before impact and the subsequent compression of the spring.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, questioning the calculations related to the speed and spring compression. There are inquiries about potential errors in the method, particularly regarding the use of gravitational force and significant figures.

Discussion Status

The conversation is ongoing, with some participants expressing confidence in the original poster's method while others suggest possible reasons for discrepancies in the answers provided to the web-based homework system. There is no clear consensus on the source of the issue, but multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the web-based homework system may have specific requirements regarding significant figures, and there is mention of potential errors in the instructor's input of the correct answers.

jamesm113
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The cable of a 3750 lb elevator in the figure below snaps when the elevator is at rest at the first floor so that the bottom is a distance d = 12.0 ft above a cushioning spring whose force constant is k = 10,000 lb/ft. A safety device clamps the guide rails, removing 1000 ft-lb of mechanical energy for each 1.00 ft that the elevator moves.
13-21.gif


(a) Find the speed of the elevator just before it hits the spring. mgh-1000h=mv^2/2 - 3750(12)-1000(12) = (3750/32)v^/2

v= 23.7318

(b) Find the distance that the spring is compressed. I got 2.8587 ft for this, but apparently it's wrong. Here's what I did:

mv^2/2 + mgx - 1000x = 1/2kx^2
(3750/32)(23.7318^2)/2 + 3750x - 1000x = 10000x^2/2
33000 + 2750x = 5000x^2
x= 2.8587
 

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jamesm113 said:
mv^2/2 + mgx - 1000x = 1/2kx^2
(3750/32)(23.7318^2)/2 + 3750x - 1000x = 10000x^2/2
33000 + 2750x = 5000x^2
x= 2.8587
Perhaps, you forgot to multiply by 'g'?
 
3750 lbs. = mg, right?
 
Your answers (and method) look OK to me. What makes you think it's wrong?
 
I use webassign, a web based homework system and I entered the answer and it said I was wrong.
 
Often those systems are picky about the number of significant figures. Does it tell you why your answer is wrong?
 
no, and this one gives an error margin of 1%. I've never had sig fig problems before.
 
jamesm113 said:
3750 lbs. = mg, right?

Ahh yes, I forgot your in pounds. Being a Brit, I'm used to working in kilos and rarely have to deal with lbs. As Doc Al says, I can't see anything wrong with your method. Does webassign allow your to enter exact solutions? In which case the exact solution would be;

[tex]x = \frac{1}{40}\left( 11 + \sqrt{10681} \right)\; m[/tex]
 
If you didn't make a significant figures mistake, then the only problem could be an incorrectly entered solution by the instructor. Perhaps he accidentally entered the negative root from the quadratic function instead of the positive root. I am convinced that it is not your mistake.
 

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